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Solutions (Chapter 12)

Solutions (Chapter 12). Taylor, Tommy, Kayla. Homogeneous mixture - mixture of two or more substances that has a uniform appearance throughout ex: salt water Heterogeneous mixture - can be separated physically and the different components are visibly distinguishable from one another

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Solutions (Chapter 12)

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  1. Solutions (Chapter 12) Taylor, Tommy, Kayla

  2. Homogeneous mixture - mixture of two or more substances that has a uniform appearance throughout ex: salt water Heterogeneous mixture - can be separated physically and the different components are visibly distinguishable from one another ex: chocolate chip cookie http://merryann12.hubpages.com/hub/Heterogeneous-and-Homogeneous-Mixtures

  3. Solute - substance being dissolved ex: Kool-Aid mix Solvent - dissolves another substance ex: water Solution - a homogeneous mixture of two or more substances. ex: Kool-Aid https://www.google.com/search?q=kool+aid&oe=utf-8&aq=t& http://www.chem4kids.com/files/matter_solution.html

  4. Visual Salt Water/ Solution Salt/ Solute Water/ Solvent google.com

  5. Suspension - a mixture in which large particles can be evenly distributed and the components will settle out ex: muddy water Colloid - a mixture in which small particles can be mixed so they remain evenly distributed without settling out ex: milk http://www.800mainstreet.com/9/0009-001-mix-solut.html

  6. Dilute - a solution containing a relatively small quantity of solute as compared with the amount of solvent ex: bleach Concentrated - a solution that contains a large amount of solute relative to the amount that could dissolve ex: alcohol http://www2.ucdsb.on.ca/tiss/stretton/chem2/acid02.html

  7. Unsaturated - the solute concentration is lower than its equilibrium solubility Saturated - a point of maximum concentration in which no more solute can be dissolved Supersaturated - increase of concentration beyond saturation http://shikagami21.blogspot.com/2007/08/saturated-unsaturated-and.html

  8. Solubility Curves Solute below the line indicates the solution is unsaturated at a certain temperature Solute above the line shows all of the solute has dissolved and is supersaturated If the amount of solute is on line then the solution is saturated Ex: At 500C and 100g, NaNo3 is saturated At 200C and 100g, NaNo3 is supersaturated At 200C and 50g, NaNo3 is unsaturated http://gcserevision101.files.wordpress.com/2009/02/solubility-curves.jpg

  9. Stirring a solution helps disperse the solute particles and bring fresh solvent into contact with the solute surface. This increases the contact between the solvent and solute. Stirring A Solution google.com

  10. Surface Area • The dissolution process occurs at the surface of the solute, so the larger the surface area of the solute the faster it dissolves. google.com

  11. Temperature of a Solution • As the temp. of a solvent increases the solvent molecules move faster. This causes collisions between the solvent and solute molecules. This helps separate solute molecules and disperse them among the solvent molecules. google.com

  12. Concentration When you increase the concentration of the solute, the rate of solution formation will also increase. They are directly related.

  13. Molarity The concentration of a solution expressed as the number of moles of solute dissolved in each liter of solution moles of solute or moles Molarity= Liters of solution L

  14. Molarity Calculations A saline (salt water) solution contains 0.70 g of NaCl per 100 ml of solution. What is its molarity? First calculate the number of moles NaCl in the solution. You must use dimensional analysis to convert from grams to moles. 0.70 g NaCl x 1 mole NaCl = 0.01 mole 58.44 g NaCl

  15. Molarity Calculations Where do the numbers come from? 0.70 g NaCl x 1 mole NaCl = 0.01 mol 58.44 g NaCl Formula mass correlates with grams in one mole Formula mass of NaCl from periodic table Grams of solute given in problem

  16. Molarity Calculations Now that you have obtained the number of moles solute, use this number in the molarity equation: M= 0.01 moles NaCl 0.1L To find this value, convert from ml (given in the problem) to liters. M= 0.1

  17. Molarity Practice 1. How many grams of CaCl2 are needed to make 125 ml of a 1.5 M solution of CaCl2? 2. How many moles of solute are present in 750 ml of a 0.10 M Na2SO4 solution?

  18. Molarity Answers 1. How many grams of CaCl2 are needed to make 125 ml of a 1.5 M solution of CaCl2? 1.5M = moles of solute .125 L .188 mol x 110.98 g CaCl2 = 20.9 g 1 mole CaCl2

  19. Molarity Answers 2. How many moles of solute are present in 750 ml of a 0.10 M Na2SO4 solution? .10 M = moles of solute = .075 moles .750 L Na2SO4

  20. Molality The number of moles of solute dissolved in each kilogram of solvent moles of solute or moles Molality= Kilograms of solvent Kg Molality is represented by "m "

  21. Example Molality Problem: What is the molality of a solution if 100 g of NaCl is dissolved into 250 g of water? First calculate the number of moles NaCl in the solution. You must use dimensional analysis to convert from grams to moles. 100g NaCl x 1 mole NaCl = 1.71 moles NaCl 58.44g NaCl

  22. Molality Calculations Where do the numbers come from? 100g NaCl x 1 mole NaCl = 1.71 mol NaCl 58.44g NaCl Formula mass correlates with grams in one mole Formula mass of NaCl from the periodic table Grams of solute given in problem

  23. Molality Calculations Now that you have obtained the number of moles solute, use this number in the molality equation: m = 1.71 moles NaCl .25 Kg m = 6.84 To find this value, convert from g (given in the problem) to Kg.

  24. Molality Practice 1. A solution was prepared by dissolving 8.55 g of sucrose (table sugar, C12H22011) in 62.5 g of water. Find the molality of this solution. 2. A solution of iodine, I2, in carbon tetrachloride,CCl4, is used when iodine is needed for certain chemical tests. How much iodine must be added to prepare a 0.240 m solution of iodine in CCl4 if 50.0 g of CCl4 is used?

  25. Molality Answers 1. A solution was prepared by dissolving 8.55 g of sucrose (table sugar, C12H22011) in 62.5 g of water. Find the molality of this solution. 8.55g C12H22011 x 1 mol C12H22011= 0.0250 mol C12H22011 342.34 g C12H22011 m = 0.0250 mol C12H22011= 0.400 m C12H22011 0.0625 kg H2O

  26. Molality Answers 2. A solution of iodine, I2, in carbon tetrachloride,CCl4, is used when iodine is needed for certain chemical tests. How much iodine must be added to prepare a 0.240 m solution of iodine in CCl4 if 50.0 g of CCl4 is used? 50.0 g CCl4 x 1 kg = 0.05 kg CCl4 1000 g CCl4 0.240 m = X X= 0.012 moles I2 0.05 kg CCl4 0.012 moles I2 x 253.8 g I2 = 3.05 g I2 1 mole I2

  27. Dilution The concentration of a solution can be decreased by adding water. M1 x V1 = M2 x V2 Dilution Equation: Molarity after dilution Volume should always be expressed in liters Original molarity Volume after dilution Original volume

  28. Dilution Calculations How much concentrated 6.0 M hydrochloric acid is needed to prepare 50.0 ml of a 1.0 M solution? First use the information given in the question in order to substitute into the dilution equation. 6.0 M x V1 = 1.0 M x 0.05 L

  29. Dilution Calculations Now, solve for the missing value algebraically. 6.0 M x V1 = 1.0 M x 0.05 L 6.0V1 = 0.05 6.0 6.0 V1 = 0.0083 L

  30. Dilution Practice 1. To what volume should 8.3 ml of 5.0 M nitric acid be diluted to prepare a 1.0 M solution? 2. To how much water should 12.5 ml of 3.0 M hydrochloric acid be added to produce a 1.0 M solution?

  31. Dilution Answers 1. To what volume should 8.3 ml of 5.0 M nitric acid be diluted to prepare a 1.0 M solution? 5.0 M x 0.0083 L = 1.0 M x V2 0.0415 = 1.0V2 1.0 1.0 V2= 0.042 L

  32. Dilution Answers 2. To how much water should 12.5 ml of 3.0 M hydrochloric acid be added to produce a 1.0 M solution? 3.0 M x 0.0125 L = 1.0 M x V2 0.0375 = 1.0V2 1.0 1.0 V2 = 0.038 L 0.038 L - 0.0125 L = 0.026 L of H2O

  33. Further Explanation 2. To how much water should 12.5 ml of 3.0 M hydrochloric acid be added to produce a 1.0 M solution? 3.0 M x 0.0125 L = 1.0 M x V2 Molarity of solute given in problem Unknown Volume of solute given in problem Final solute molarity given in problem

  34. Further Explanation 2. To how much water should 12.5 ml of 3.0 M hydrochloric acid be added to produce a 1.0 M solution? 3.0 M x 0.0125 L = 1.0 M x V2 0.0375 = 1.0V2 1.0 1.0 V2 = 0.038 L Solve algebraically

  35. Further Explanation 3.0 M x 0.0125 L = 1.0 M x V2 0.0375 = 1.0V2 1.0 1.0 V2 = 0.038 L 0.038 L - 0.0125 L = 0.026 L of H2O Subtract the final volume and the original volume

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